Question

Consider the nonhomogeneous differential

Answer 1

Answer:

$c_1e^{3x}+c_2e^{-x}+\frac{1}{2} xe^{3x}$

Step-by-step explanation:

For homogeneous solution:

$y''-2y'-3=0\\r^2-2r-3=0\\(r-3)(r+1)=0$

So since roots are r = 3 and r = -1, $y_h=c_1e^{3x}+c_2e^{-x}$

Since we are given $2e^{3x}$, we will use undetermined coefficients. However, here the trick is we have $c_1e^{3x}$ in homogeneous solution. So in particular solution, as undetermined coefficients, we will use $Axe^{3x}$ instead of $Ae^{3x}$.

Hence,

$Y = Axe^{3x}\\Y' = Ae^{3x} + 3Axe^{3x}\\Y'' = 3Ae^{3x} + 3Ae^{3x} + 9Axe^{3x} = 6Ae^{3x} + 9Axe^{3x}$

So,

$6Ae^{3x}+9Axe^{3x}-2Ae^{3x}-6Axe^{3x}-3Axe^{3x}=2e^{3x}\\4Ae^{3x}=2e^{3x}\\A=\frac{1}{2}$

Hence, $y_p = \frac{1}{2} xe^{3x}$

General solution is:

$y=y_h+y_p=c_1e^{3x}+c_2e^{-x}+\frac{1}{2} xe^{3x}$